# Find the magnetic energy

1. Jan 27, 2006

### mmh37

Hi everyone,

this is a hard question (at least for me):

For a coaxial cable (description see below) find the magnetic energy stored in a cylindrical annulus of unit length, raduius r and thickness dr, in the regions 0<r<a and a<r<b.

Hence by integration find the magnetic energy stored in unit length of the cable and deduce the self-inductance of the cable per unit length

(you may assume that the thickness of the outer cylindrical conductor is is very small so that you can ignore the magnetic energy stored in the region b<r<c).

this is my solution/attempt:

1. magentic Energy

$$W = \frac {L*I^2} {2} = (...) = \frac {B^2} {2*mu} * Volume$$

therefore $$W/l = \frac {B^2} {2*mu} * Area$$ where the area is pi* r^2

0<r<a

$$B (r) = \frac {mu *l*r} {2*pi*a^2}$$

so substituting in energy expression derived above gives:

$$W = \frac {mu *l^2*r^4} {8*pi*a^4} dr$$

a<r<b

$$B = \frac {mu *l} {2*pi*r}$$

so $$W = \frac {mu *l^2} {8*pi}dr$$

now, to find the total energy I integrated the first expression from 0 to a and added the second, which I integrated from a to b.

The self-inductance would then be esaily calculable by W = .5* L*I^2

However, this gives the wrong result and I am 99% certain that there is no calculation error.

Can anyone see the mistake?

Thanks so much!

____________________________________________________________
Description of the coaxial cable:

a coaxial cable consists of a solid inner cylindrical conductor of radius a and an outer cylindrical conductor of inner and outer radius b and c. Distributed currents of equal magnitude I flow in opposite directions in the two conductors. expressions for the magnetic flux density B(r) for each of these regions are

1) 0<r <a

$$B = \frac {mu *l*r} {2*pi*a^2}$$

where mu is the permeability of free space

2) a<r<b

$$B = \frac {mu *l} {2*pi*r}$$

3) b<r<c

$$B = \frac {mu *l*(c^2-r^2)} {2*pi*r*(c^2-b^2)}$$

Last edited: Jan 27, 2006